Keywords Search
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19868 Accepted Submission(s): 6669
Problem Description
In the modern time, Search engine came into the life of everybody like Google, Baidu, etc.
Wiskey also wants to bring this feature to his image retrieval system.
Every image have a long description, when users type some keywords to find the image, the system will match the keywords with description of image and show the image which the most keywords be matched.
To simplify the problem, giving you a description of image, and some keywords, you should tell me how many keywords will be match.
Wiskey also wants to bring this feature to his image retrieval system.
Every image have a long description, when users type some keywords to find the image, the system will match the keywords with description of image and show the image which the most keywords be matched.
To simplify the problem, giving you a description of image, and some keywords, you should tell me how many keywords will be match.
Input
First line will contain one integer means how many cases will follow by.
Each case will contain two integers N means the number of keywords and N keywords follow. (N <= 10000)
Each keyword will only contains characters 'a'-'z', and the length will be not longer than 50.
The last line is the description, and the length will be not longer than 1000000.
Each case will contain two integers N means the number of keywords and N keywords follow. (N <= 10000)
Each keyword will only contains characters 'a'-'z', and the length will be not longer than 50.
The last line is the description, and the length will be not longer than 1000000.
Output
Print how many keywords are contained in the description.
Sample Input
1 5 she he say shr her yasherhs
Sample Output
3
Author
Wiskey
Recommend
lcy
题意：给n个单词和一个字符串，询问该字符串中有多少个给出的单词。
code:
1 #include<iostream> 2 using namespace std; 3 4 struct node 5 { 6 node *fail; 7 node *next[26]; 8 int count; 9 node() 10 { 11 fail=NULL; 12 count=0; 13 memset(next,NULL,sizeof(next)); 14 } 15 }*q[500001]; 16 17 char keyword[55]; 18 char str[1000010]; 19 20 void insert(char *str,node *root) 21 { 22 node *p=root; 23 int i=0; 24 while(str[i]) 25 { 26 int index=str[i]-'a'; 27 if(p->next[index]==NULL) 28 p->next[index]=new node(); 29 p=p->next[index]; 30 i++; 31 } 32 p->count++; 33 } 34 35 void acbuild(node *root) //构建失败指针 36 { 37 int i; 38 int head=0,tail=0; 39 root->fail=NULL; 40 q[tail++]=root; 41 while(head!=tail) 42 { 43 node *temp=q[head++]; 44 node *p=NULL; 45 for(i=0;i<26;i++) 46 { 47 if(temp->next[i]!=NULL) 48 { 49 if(temp==root) 50 temp->next[i]->fail=root; 51 else 52 { 53 p=temp->fail; 54 while(p!=NULL) 55 { 56 if(p->next[i]!=NULL) 57 { 58 temp->next[i]->fail=p->next[i]; 59 break; 60 } 61 p=p->fail; 62 } 63 if(p==NULL) 64 temp->next[i]->fail=root; 65 } 66 q[tail++]=temp->next[i]; 67 } 68 } 69 } 70 } 71 72 int query(node *root) 73 { 74 int i=0,cnt=0; 75 int index; 76 node *p=root; 77 while(str[i]) 78 { 79 index=str[i]-'a'; 80 while(p!=root&&p->next[index]==NULL) 81 p=p->fail; 82 p=p->next[index]; 83 p=(p==NULL)?root:p; 84 node *temp=p; 85 while(temp!=root&&temp->count!=-1) 86 { 87 cnt+=temp->count; 88 temp->count=-1; 89 temp=temp->fail; 90 } 91 i++; 92 } 93 return cnt; 94 } 95 96 int main() 97 { 98 int n,t; 99 scanf("%d",&t); 100 while(t--) 101 { 102 node *root=new node(); 103 scanf("%d",&n); 104 getchar(); 105 for(int i=0;i<n;i++) 106 { 107 gets(keyword); 108 insert(keyword,root); 109 } 110 acbuild(root); 111 scanf("%s",str); 112 printf("%d\n",query(root)); 113 } 114 return 0; 115 } 116 /* 117 10 118 3 119 a 120 bb 121 ca 122 aabbcca 123 */